Question:

<p><strong>Step 1: Add all rows to Row 1</strong></p><p>$$R_1 \to R_1 + R_2 + R_3$$</p><p>First row becomes: $(a-b-c+2b+2c, 2a+b-c-a+2c, 2a+2b+c-a-b)$</p><p>Simplifying: $(a+b+c, a+b+c, a+b+c)$</p><p>Factor out $(a+b+c)$:</p><p>$$\begin{vmatrix} 1 & 1 & 1 \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix} \cdot (a+b+c)$$</p><p><strong>Step 2: Subtract Column 1 from Columns 2 and 3</strong></p><p>$$C_2 \to C_2 - C_1, \quad C_3 \to C_3 - C_1$$</p><p>$$\begin{vmatrix} 1 & 0 & 0 \\ 2b & b-c-a-2b & 2b-2b \\ 2c & 2c-2c & c-a-b-2c \end{vmatrix}$$</p><p>$$= \begin{vmatrix} 1 & 0 & 0 \\ 2b & -b-c-a & 0 \\ 2c & 0 & -a-b-c \end{vmatrix}$$</p><p><strong>Step 3: Expand along Row 1</strong></p><p>$$\det = 1 \cdot \begin{vmatrix} -b-c-a & 0 \\ 0 & -a-b-c \end{vmatrix}$$</p><p>$$= (-a-b-c) \cdot (-a-b-c) = (a+b+c)^2$$</p><p><strong>Step 4: Combine with the factor from Step 1</strong></p><p>$$\text{Original determinant} = (a+b+c) \cdot (a+b+c)^2 = (a+b+c)(a+b+c)^2$$</p><p><strong>Step 5: Match with given form</strong></p><p>Given: $\det = (a+b+c)(x+a+b+c)^2$</p><p>Comparing: $(a+b+c)(a+b+c)^2 = (a+b+c)(x+a+b+c)^2$</p><p>Therefore: $(a+b+c)^2 = (x+a+b+c)^2$</p><p>This gives: $a+b+c = x+a+b+c$ or $a+b+c = -(x+a+b+c)$</p><p>From the first: $x = 0$ (excluded by the problem)</p><p>From the second: $a+b+c = -x-a-b-c \implies 2(a+b+c) = -x \implies x = -2(a+b+c)$</p><p><strong>∴ Answer: D</strong></p>